TEST IV, Exercise 5. Correction

We consider the 2 following algebraic numbers:
α=(((i+√2))^(1/7))+(((i+√2))^(3/7))+(((i+√2))^(5/7))+1
and
β=(((√2+(3^(1/3))))^2)+√(1+(3^(1/3)))
The aim of this exercise is to compute a vanishing polynomial (Subscript[p, α]) of α and a vanishing polynomial (Subscript[p, β]) of β.

Question 1
Let (Subscript[α, 1])= i + √2. Compute a vanishing polynomial (Subscript[p, (Subscript[α, 1])])of (Subscript[α, 1]).

a1=I+Sqrt[2]

Expand[a1^2]

Expand[(a1^2 -1)^2]

pa1=(x^2-1)^2+8

Indeed, we have pa1(a1)=0 !! So pa1 is a vanishing polynomial for a1.

Deduce a vanishing polynomial (Subscript[p, (Subscript[α, 2])])of (Subscript[α, 2])=((Subscript[α, 1])^(1/7))

a2=a1^(1/7)

We remark that pa1(a2^7)=pa1(a1)=0, so pa2(x)=pa1(x^7) is  a vanishing polynomial of a2

pa2=pa1/.x->x^7

Expand[pa2]

Question 2
Show that there exists a polynomial h such that α=h((Subscript[α, 2])). Use the resultant (Lecture VI, Part II, 2) to compute (Subscript[p, α])

We remark that α=((Subscript[α, 2])^0)+(Subscript[α, 2])+((Subscript[α, 2])^3)+((Subscript[α, 2])^5), so h=1+x+x^3 + x^5

h=1+x+x^3+x^5

pa=Resultant[y-h,pa2,x]

Question 3
Compute a vanishing polynomial of 1+(3^(1/3)).

b0=1+3^(1/3)

Expand[(b0-1)^3]

pb0=(x-1)^3-3

pb0/.x->b0

We see that pb0 is a vanishing polynomial for b0

Deduce one for √(1+(3^(1/3)))

b02=Sqrt[b0]

pb0(b02^2)=pb0(b0)=0, so pb0(x^2) is a vanishing polynomial for b02

pb02=pb0/.x->x^2

Expand[%]

Question 4
Use the resultant to compute a vanishing polynomial of (Subscript[β, 1])=√2 + (3^(1/3)).

b1=Sqrt[2]+(3^((1/3)))

x^2-2 is a vanishing polynomial for √2 
x^3-3 is a vanishing polynomial for 3^(1/3)

pb1=Resultant[(y-x)^2-2,x^3-3,x]

Expand[pb1/.y->b1]

Deduce one for (Subscript[β, 2])=((Subscript[β, 1])^2)

b2=b1^2

pb2=Resultant[y-x^2,pb1/.y->x,x]

Expand[pb2/.y->b2]

We see that pb2 is a vanishing polynomial for b2

Question 5
Use the resultant to compute (Subscript[p, β])

We see that β=b02+b2

pb=Resultant[pb2/.y->y-x,pb02,x]

β = (Sqrt[2] + 3^(1/3))^2 + Sqrt[1 + 3^(1/3)]

Expand[pb/.y->β]

YEAH ! It works

Question 6
Use the resultant to compute (Subscript[p, αβ])

Expand[pa2]

Pa=Expand[x^28 ( pa2/.x->(y/x))]

Resultant[Pa,pb/.y->x,x]